Equality of two Determinants (transformation)

Question

$det\begin{pmatrix} -\lambda & 1 & 1 & 1\\1 & -\lambda & 1 & 1\\1 & 1 & -\lambda & 1\\1 & 1 & 1 & -\lambda\\\end{pmatrix} = det\begin{pmatrix}-\lambda+3 & 1 & 1 & 1\\-\lambda + 3 & -\lambda & 1 & 1\\-\lambda+3 & 1 & -\lambda & 1\\-\lambda+3 & 1 & 1 & -\lambda\\\end{pmatrix}$

I fail to see how this transformation is justified. In the solution to my practice exercises, there is no explanation to this step. Can someone enlighten me?

Answer 1

You just have to add the columns #2, #3 and #4 to the first.

Elementary row- and/or column-operations on your matrix do not change the value of its determinant.

Answer 2

Adding all the three last columns to the first one doesn't change the value of the determinant.

Answer 3

Call the first matrix $A$ and the second $B$. Note that $B=AE$ where $$ E= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{pmatrix} $$ But now $\det(B)=\det(AE)=\det(A)\det(E)=\det(A)\cdot 1=\det(A)$.